\(\displaystyle 4*\int_{0}^{1}\frac{1}{1+x^2}dx = 4tan^{-1}(x)\) 

Evaluated between 0 and 1.

\(\displaystyle =4*tan^{-1}(1)-4*tan^{-1}(0)=4*\frac{\pi }{4}-4*0=\pi\)

When we look at the figure we are trying to integrate, we see that it is actually a quarter circle.

Using the rules we learned in geometry, we know that the area of a circle is \(\displaystyle \pi r^2\). As such, we can take the radius and square it and then multiply it by \(\displaystyle \pi\). Our radius is found by taking the square root of the constant under the radical.

So now that we have the area of the whole circle, we need to divide it by 4.

\(\displaystyle 4\pi\div4=\pi\)

When we look at the shape of the graph we notice that it is a triangle with a height of 3 and a base of 3.

So the area under the curve would be the same as the area of this triangle.

From geometry, from geometry we know that the area of a triangle is

\(\displaystyle \frac{1}{2}\times (\text{Base})\times (\text{Height})\).

So our answer would be 

\(\displaystyle \frac{1}{2}\times 3\times 3=\frac{9}{2}\)

The first thing we must notice is that this is an odd function Given the properties of odd functions we know that it is a reflection over the origin.

This means that the same amount is above the axis as is below.

This means that the overall area is zero.

First we see that there is no special cases, so we can just use the power rule to integrate.

We add one to each power and then divide by the power so we end up with 

\(\displaystyle 2x^{3}+2x^{2}\).

After that we plug in our top bound and find our value.

Then we plug in our bottom bound and find our value. We take these two values and subtract the top value from the bottom value.

\(\displaystyle \\2(4)^{3}+2(4)^{2}-(2(2)^{3}+2(2)^{2}) \\2(64)+2(16)-(2(8)+2(4)) \\128+32-(16+8) \\160-24 \\136\)

It is important to recognize this as it is a common and important integral, it is the natural log of x or more appropriately \(\displaystyle ln\left | x\right |\) . What we should look for when seeing this is 1/x or some variation of this.

When the integrals get harder and we start to do u-substitution, we will look for the derivative over the original.

But since in this problem, we know the integral will be \(\displaystyle ln\left | x\right |\), we just plug in the top value and get 1.

Plug in the bottom value and get 0.

Then subtract the two values to get 1.

This is another important integral to remember and that is e^x is always e^x.

Whether we integrate or derive, e^x will stay the same.

So it becomes important to just plug in out top bound, we get 5.

Plug in our bottom bound and we get 1.

So our answer is 5-1 which is 4.

For this problem we will use u-substitution. U-substitution can be thought of the reverse chain rule. We see we have a function we would have used the chain rule on and then we back track it. Here we can see that we have \(\displaystyle x^{2}\) and the neighbor down the block \(\displaystyle x\). This helps identify what the u for our substitution will be. so we take the stuff we have inside the parenthesis and we set that equal to u. So..
\(\displaystyle u=x^2+1\)

We derive that so we can find our dx.

\(\displaystyle du=2xdx\)

Solve for dx.

\(\displaystyle dx = \frac{du}{2x}\)

Now we subsitute it back in.

\(\displaystyle \int_{0}^{1}18x(u)^8\frac{du}{2x}\)

Simplify.

\(\displaystyle \int_{0}^{1}9u^8du\)

Integrate.

\(\displaystyle u^9 from 0 to 1\)

Plug in original.

\(\displaystyle (x^2+1)^9 from 0 to 1\)

Plug in bounds.
512-1

Solve.

511

First things is first, we identify what our u is. In this case, we want to look at our denominator. We see that we have \(\displaystyle x^2\) and in the numerator we have \(\displaystyle x\). We can see they are next door neighbors. So our u is...

\(\displaystyle u=x^2+1\)

Next we derive.

\(\displaystyle du=2xdx\)

Solve for dx.

\(\displaystyle dx=\frac{du}{2x}\)

Substitute back in.

\(\displaystyle \int_{0}^{1}\frac{6x}{u}\frac{du}{2x}\)

Simplify.

\(\displaystyle \int_{0}^{1}\frac{3}{u}du\)

Integrate.

\(\displaystyle 3ln|u||_{0}^{1}\)

Plug in the original.

\(\displaystyle 3ln|x^2+1||_{0}^{1}\)

Plug in top bound and bottom pound.

\(\displaystyle 3ln|2|-3ln|1|=ln|8|\)

First we must define what equation we will use to solve this integral. The equation to solve integration by parts is.

\(\displaystyle \int{udv}=uv-\int{vdu}\)

So we must define what our u and dv are. U is generally the simpler of the two terms or the natural log. Dv is generally the more complicated of the two. We will take the derivative of U and find the integral of dv.

So

\(\displaystyle u=x\)

Derive.

\(\displaystyle du=dx\)

and

\(\displaystyle dv=e^xdx\)

Integrate.

\(\displaystyle v=e^x\)

Now that we have all the parts, it is just plug and work out.

\(\displaystyle \int_{0}^{1}{xe^xdx}=xe^x-\int_0^1{e^xdx}\)

Take the second part and integrate.

\(\displaystyle xe^x-e^x|_{0}^{1}\)

Plug in top bound and bottom bound.

\(\displaystyle 1e^1-e^1-(0e^0-e^0)\)

\(\displaystyle 0-(-1)=1\)